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Back in the 7th grade, our science teacher Mr. Dabney made a bet with us the first day of school. There were about 35 students in the class, and he wagered 35 boxes of donuts there were at least 2 of us that shared the same birthday (he promised he didn’t know this information beforehand).
We took him up on the bet, confident he was making a foolish mistake, but as Mr. Dabney made his way around the room asking our birthdays, a match popped up just as he got to the last couple of students.
So the question is: How many people do you need to have in a room so that there is a greater than 50% chance that 2 of them will share the same birthday?
How many people do you need to have a 97% chance that 2 will share the same birthday?
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Not entirely sure, but I think it’s 14 an 19, respectively.
this reminds me of something interesting that happened to me last month…. i work in a restaurant, and therefore often check IDs when young people order alcohol. i checked the ID of a young lady and said ‘hm, you have the same birthday as my mom’–then i checked the guys ID and was suprised to see he shared the same birthday as my sister! i figured the odds of one birthday being shared were 1 in 365, (365.25 if you want to be really accurate) and the odds of 2 birthdays matching would be 365.25X365.25 or 1 in 133,407. kinda random. i guess this would be the same odds as 3 strangers sharing a single birthday….
as for the problem at hand, im not sure. im not quite sure how to set up such an equation. it seems to me that to ensure that there is a 50-50 chance of 2 people sharing a birthday, at least half the days would have to be accounted for, but if you had 182 people, it seems better than a long shot that 2 would match….and jr. high teachers wouldnt be placing such wagers. ill have to think about this. way to make me lose sleep dr. kaplan!!
great question.
X=average number of days in a month
you would have to have at least x amount of people in a room to have atleast a 50% chance. so the equation would be 2 out of X. or 1 out of X divided by two. to have atleast a 97% chance, you would have to have twice as many kids as X. so this equation would be 2 out of two times X, or 1 out of X.
^thats just what i think.
Well its a matter of statistics. since the first birthday doesnt matter, we can use 1/1 as the fraction for that, and 1/365 since it has to be on the same day as the first for the second.
so 365/2 is the number of people in the same room youd need to have to have a 50% chance.
and .97 times 365 for a 97% chance.
^^sorry dad, but the 35 kids in that room is much less than 365/2
http://en.wikipedia.org/wiki/Birthday_paradox
The chance that someone is born on any day in a year is: 1/365
Since the birthday event is independent of any other b-day event…the chance that any 2 people have the chance of being born the same day is:
(1/365)*(1/365)….and yet again one has to consider the chance that this happens on any one day choosing 1 day out of 365 = 365 possible dates and in this class there are 35 potential candidates meaning that there are 2 out of 35 possible combinations….
the solution to the possibility of 2 people sharing the same B-day should thus be:365* 1190*(1/365)*(1/365)
where
365= results from 365 chose 1 day
1190= results from 35 chose 2 candidates
bigger 50% –> 15 candidates
bigger 97%–> 20 candidates
[...] If I were a teacher, I’d offer my class a morning of free donuts if anyone in the class figured this puzzle out : ) [...]